TITLE: Transforming trigonometric identities
QUESTION [5 upvotes]: The problem goes like this:
If
$$N=2\sec^4x-3\sec^2x+2=\frac{\cos^2x}{\cos^2y}$$
Calculate the equivalent of
$$M=2\tan^4x+3\tan^2x+2$$
The alternaties I have are:
$$\frac{\tan^2x}{\tan^2y},\mbox{ }\frac{\tan^2y}{\tan^2x},\mbox{ }\frac{\tan^2y}{\sec^2x},\mbox{ }\frac{\sec^2y}{\tan^2x},\mbox{ }\frac{\sec^2x}{\tan^2y}$$
The first thing I tried was to "build" the value of $N$ and then use it for $M$
$$\sec^2x=1+\tan^2x\\
\sec^4x=1+2\tan^2x+\tan^4x\\
2\sec^4x=2+4\tan^2x+2\tan^4x\\
-3\sec^2x=-3-3\tan^2x\\
2\sec^4x-3\sec^2x+2=1+\tan^2x+2\tan^4x\\
\frac{\cos^2x}{\cos^2y}=1+\tan^2x+2\tan^4x\\
\frac{\cos^2x}{\cos^2y}+2\tan^2x+1=2+3\tan^2x+2\tan^4x\\
\frac{\cos^2x}{\cos^2y}+2\tan^2x+1=M$$
But then I can't transform the final equation to one of the alternatives, even after trying a massive substition of $\cos^2y$ it didn't helped too much. Any hints or ideas are greatly appreciated.

REPLY [3 votes]: (I do not like this solution, and do not have a good explanation for the observation, which is based on the potential answers. There is no hint that I can give, which doesn't give away the entire game.)
Observe that $$M \sin^2x - N = - \cos^2x ,$$
which you can check in Wolfram (see alternate forms).
Hence, the answer is __ (fill in the blank yourself).

The issue that Cameron was having arose, because we have to divide by $\sin^2 x$ (which is 0 when $x=0$) in determining the answer.